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need a simple gadget - an AC current detector audible alarm

Discussion in 'Misc Electronics' started by Steve Grubb, Sep 20, 2004.

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  1. Steve Grubb

    Steve Grubb Guest

    Hi, I'm looking for something with AC male on one end, and AC female
    on the other, that can plug an AC device (sump pump) into, that will
    emit an audible tone when the device is running and consuming current.
    Kind of like an audible pilot light, or an electric meter that
    sings.. I want this so I can be aware upstairs when the sump pump runs
    in the cellar, since occasionally it sticks on. This would also alert
    us to a plumbing problem. It's a dirt cellar.

    It seems like would be a simple device, but I don't know what it would
    be called.

    Thanks for any ideas. -Steve
  2. CWatters

    CWatters Guest

    Perhaps you could just wire a mains lamp or a mains powered radio across the
    pump motor (tune radio to 24 hour station!)
  3. I would use a big, heavy full-wave bridge rectifier, something rated
    for more amps than the sump pump will draw. I've seen 25A bridge
    rectifiers available for several dollar each. Or I might try a bridge
    rectifier out of an old scrapped AT power supply if it's rated at 6A or
    more. Depends on the pump.

    Then I'd get a pigtail with a plug on it and a socket and insert the
    bridge's two "~" AC leads in series with the hot AC wire between the
    plug and socket.

    Then I'd put a 1 ohm, ten watt resistor across the bridge's + and -
    leads. With the pump plugged into it, I'd measure the voltage drop
    across the resistor. Be Careful around AC, it can be deadly! The idea
    here is to get the voltage drop across the resistor to be enough to make
    a buzzer sound. It might take only a 1 ohm resistor if the pump draws a
    few amps, giving a few volts drop across the resistor. Or if the pump
    is lower current, then it might take 2 or 5 ohms. But the resistor has
    to be a power resistor to handle the current. With a 1 ohm resistor, 3
    volts drop times 3 amps gives 9 watts of dissipation, so the resistor
    will get hot. Of course, a piezo buzzer goes across the resistor, in
    the proper polarity.

    Make sure everything is well insulated, since the components have the AC
    line voltage on them. I would put all of it inside a metal electrical
    outlet box, wide enough for both an outlet and a switch. Instead of the
    switch, leave the hole for the switch handle empty and mount the buzzer
    so that it's below and the sound can come out the hole.

    Remember all this has to reliable enough to never prevent the pump from
    working. It could be messy if it doesn't do its job.
  4. Problem is that the AC line going to the motor is usually hot all the
    time. The soundmaker would have to be wired _after_ the float switch,
    which in the sump pumps I've seen, isn't easily accessible because it
    has to be insulated from the water.
  5. CWatters

    CWatters Guest

    So you really need to detect the change in current when the motor switches
    on rather then the voltage (which is constant).
  6. CWatters

    CWatters Guest

    I saw a gadget advertised in a PC mag recently... It looked like a 4 way
    extension block with one special socket into which you plug your PC. It
    detected when you switched off your PC (by measuring the current drop) and
    then 10 seconds later it automatically switched off power to the other
    devices. Perhaps this would do the job.
  7. Bill Bowden

    Bill Bowden Guest

    Well, you could use a piezo buzzer and a
    few other parts. One resistor, one diode and
    one capacitor. The piezo buzzer should be
    the type that operates on a low DC voltage
    and doesn't require a driver.

    Just wire the circuit in parallel with the AC
    connections to the pump. You can adjust the 16K
    resistor for more or less volume, but don't take
    it too low or the thing might fry. The diode needs
    to be a 200 volt variety or greater for 120 AC
    use. A 1N4003 or 1N4004 should work.

    16K Diode
    Pump ---\/\/\------->|------+-------+
    AC + | |
    --- Piezo buzzer
    10uF --- |
    | |
    | |
    Pump -----------------------+-------+

  8. CWatters

    CWatters Guest

    He said that wasn't possible because the float switch and motor connections
    are sealed inside the pump. He only has access to power going in so he has
    to measure the current somehow.
  9. John Fields

    John Fields Guest

    +-------------|~ +|------+-----+
    | +-----|~ -|--+ |+ |
    | | +----+ | [C] [BUZZER]
    MAINS>---+--[R]--+----+ | | |
    | +---+-----+

    Select the lowest voltage buzzer (beeper, Sonalert, etc.) you can
    find, and choose R so that the current drawn by the pump motor will
    drop the voltage needed by the buzzer plus 1.4V for the diodes in the
    bridge. That is:

    R = (Vbuzzer + 1.4V) / I motor

    The capacitor, C, will be:

    C = IdT/dV

    where C = the capacitance in Farads
    I = the steady-state current ion the buzzer
    dt = the period of the rectified waveform
    dv = the permissible ripple voltage across the buzzer

    For example, lets say you have a pump which draws 1A and you've chosen
    a 6 volt Sonalert which needs 100mA.

    For R,

    R = (Vb + 1.4V) / 1A = 7.4/1 = 7.4 ohms.

    7.5 ohms is a standard value, and then, since the resistor will drop
    7.5V at 1A, it will dissipate 7.5 watts, so you'll need to get
    something like a 15 or 20 watt resistor.
    A 20 watter won't get as hot as a 10 watter, so that would be the way
    to go, IMO. A good idea would be to use a wirewound metal-cased 25
    watt resistor which you could bolt onto the inside of a small metal
    minibox which you could use to house the rest of the components.
    Restricting all of the electrical connections to the inside of the
    enclosure and making sure there were no electrical connections made to
    the [metal] housing itself would go a long way to reducing the danger
    of shock, since you're using the unisolated mains to run the buzzer.

    For C, assuming you can tolerate 1/2 a volt of ripple:

    C = Idt / dV = 0.1A*0.0083s/0.5V = .00166µF ~ 1700µF

    Since the tolerance on the capacitance of cheap electrolytics is
    often rated at -20 to +80%, you'd need about 2200µF to make sure you
    got 1700µF, and the cap would need to be rated for at least 6V. 12V
    wouldn't hurt.

    Pretty much any old bridge would work, or you could make up something
    out of 1N4000s, so your circuit would wind up looking like this:

    +----------------|~ +|-------+-------+
    | +-----|~ -|--+ |+ |
    | | +----+ | [2200] [BUZZER]
    MAINS>---+--[7.5R]--+----+ | | |
    | +----+-------+

    If you wanted to cut down on the dissipation you could substitute a
    current transformer for the 7.5 ohm resistor, but that's another

    Whatever you decide to do be careful, since mains voltage can kill.
  10. Tweetldee

    Tweetldee Guest

    Take a look at the circuit at,
    "AC Line Current Detector". You can build it with common components.
    Should do your task nicely.

    Dave M
    MasonDG44 at comcast dot net (Just subsitute the appropriate characters in
    the address)

    Never take a laxative and a sleeping pill at the same time!!
  11. You omitted one very important specification. The 16k resistor must be
    able to handle the power, and IMNSHO, it should be able to handle the
    amount of power the whole AC line would put on it. In this case that's
    120VAC * 120VAC / 16000, which is 0.9 watts. So call it an even watt.
    And it should be flameproof.

  12. This solution is simple and cheap. The only drawback is heat. The heat
    production can be lowered by using schottky- or germanium diodes in the
    rectifier bridge and finding a buzzer that requires a low voltage. The lower
    the better.

    A low voltage AC buzzer should be even better as you don't need a rectifier
    and a capacitor to drive it.

    Another possibility is replacing the DC buzzer by a solid state relay (SSR).
    The common ones can be driven by a voltage as low as 3V. The switch can
    drive a mains powered buzzer. A simple electrical doorbel for instance
    (through its transformer of course.) A drawback may be the cost. A simple
    SSR will do about $10,--.

    As for producing the less heat possible, a current transformer is the best
    solution I can imagine. But even if you can find one it will be expensive.
    You can make one yourself by disassembling a mains to low voltage
    transformer. An old wallwart will contain one. (Not the newer ones. They are
    SMPSs.) Remove the mains coil and replace it by some turns of thick wire,
    thick enough to handle the current to your pump. The series resistor has to
    be replaced by that coil. To find out the number of turns required you will
    have to do some experimenting. Start with let's say ten turns and increase
    this number until the secundary voltage is high enough to drive the buzzer.
    (The voltage/turn will be more or less constant. So if ten turns produce 2V
    you will require twentyfive turns to produce 5V.)

    The most important and may be most expensive part is the enclosure. To stay
    on the safe side you need a solid one. Make sure that any metal part that
    can be touched from the outside has been firmly tied to the protective
    ground. Otherwise you may have build a killing gadget.

    petrus bitbyter
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